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CCNA Exam Prep Question of the Week

You have been assigned the 172.17.8.0/21 address space for your office. You have a few LANs with a maximum of 70 device addresses, and some point-to-point WANs. You use the zero subnet. You use the all 1s LAN subnet to further subnet for the WANs. What is the address of the second WAN subnet and what is the broadcast address on that subnet?

172.17.15.144; 172.17.15.147

172.17.15.4; 172.17.15.7

172.17.7.4; 172.17.7.7

172.17.16.12; 172.17.16.15

172.17.15.132; 172.17.15.135

The correct answer is 5.

For 70 device addresses, 7 host bits are needed so a mask of /25 should be used for the LANs; this means 4 bits of further subnetting for the LANs from the given /21. The all 1s subnet has bits 1111. A mask of /30 should be used on the point-to-point WANs; this means 5 bits of further subnetting. The second WAN subnet has bits 00001. The address of this subnet is 172.17.00001111.10000100. This is 172.17.15.132; the broadcast address is 172.17.15.135.

4 comments

When doing VLSM you always start with the subnets that need the MOST device addresses; in this case it is the LANs with a max of 70 addresses. For 70 device addresses, 7 host bits are needed (because 27 – 2 = 126 hosts, which is more than the 70 needed). With 7 host bits subtracted from the 32 bits in an address, 25 bits are left for network and subnet, so a mask of /25 should be used for the LANs. The given mask is /21, so this means that 4 additional bits of subnetting is needed for the LANs. These 4 additional bits can create 24 = 16 subnets. The “all 1s” LAN is the address with all 1s in these four bits.
The given address is (with the first 21 bits shaded):
172.17.8.0/21 = 172.17.00001000.00000000

A mask of /30 should be used on the point-to-point WANs because only 2 host addresses are needed so only 2 host bits are needed (because 22 – 2 = 2 hosts). From the /25 for the LANs there are therefore 5 bits of further subnetting for the WANs.

The first WAN subnet address would have all 0s in these 5 bits and would be:
172.17.15.128/30 = 172.17.00001111.10000000
The second WAN subnet address would have all 00001 in these 5 bits and would be:
172.17.15.132/30 = 172.17.00001111.10000100
The broadcast on this second WAN would have “all of the host bits” set to “1”… there are only 2 host bits, so set them both to 1:
172.17.15.135/30 = 172.17.00001111.10000111

More explanation is needed for me. I’m getting different results when I subnet. The way the question is worded, it seems that you subnet for the 70 addresses (/25) for the LANs and then use the all ones subnet for the WAN.
However it seems like you start with the /21 address and use the all ones from it.

Shouldn’t you start from 172.17.16.0/25 and then find your WAN subnets? Also the sentence “These 4 additional bits can create 24 = 16 subnets.” makes no sense to me.

The original author of the post is not available, so we’ve reached out to another Cisco subject matter expert to provide additional context. See below for the response:

This is a variable subnet masking exercise. We start with a subnet that has a /21 mask and we need subnet that to fill the needs for 70 hosts per LAN subnet. 7 host bits give us 126 possible hosts (27 = 128 – 2, 6 bits would only give us 62 possible hosts), which would give us a /25 mask. Since we’ve repurposed 4 bits for additional subneting, we can now create 8 subnets with a /25 mask from the original /21 subnet. Assuming that the last possible subnet of this size is to be used for the WAN links, then we would have the following LAN subnets:
172.16.0001 LLLL.LHHH HHHH (where L are bits to subnet for LANs and H are host bits)
1st subnet 172.16.8.128/25
2nd subnet 172.16.9.0/25
3rd subnet 172.16.9.128/25
4th subnet 172.16.10.0/25
…
14th subnet 172.16.14.128/25
15th subnet 172.16.15.0/25
16th subnet 172.16.15.128/25

Per the question, “use the all 1s LAN subnet to further subnet for the WANs.” We will use the 16th subnet for the WAN segments with a /30 mask. This would give us 2 hosts per segment. So now the address will be 172.16.15.1WWW WWHH (where W are the bits used to subnet for WAN and H are host bits). The first subnet of that size would be 172.16.15.128/30. The second would be 172.16.15.132/30. The broadcast for a subnet is when all host bits are ones, so the last two bits in this case, so the broadcast for this network would be 172.16.15.135. The only two valid address would be 172.16.15.133 and 172.16.15.134.

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